position vs time
position (m)
time (sec)
1. what is the position at t = 11s?
2. what distance was traveled from t = 3s to 5s?
3. during what time intervals was there no motion?
4. during which time interval was motion the slowest?
5. what distance was traveled from t = 17s to 20s?

Question

Accepted Answer

### 1. What is the position at $ t = 11 \, \text{s} $?
# Explanation:
## Step 1: Analyze the graph
Look at the "Position vs Time" graph. Find the point corresponding to $ t = 11 \, \text{s} $ on the time axis (x - axis) and then determine the position (y - axis) at that time. From the graph, the line from $ t = 8 \, \text{s} $ to $ t = 15 \, \text{s} $ has a slope, and at $ t = 11 \, \text{s} $, we can see the position.
## Step 2: Read the position
By looking at the graph, when $ t = 11 \, \text{s} $, the position is $ 6 \, \text{m} $ (assuming the grid lines and the graph's scale: the y - axis has markings, and at $ t = 11 $, the position value from the line is 6 m).
# Answer:
The position at $ t = 11 \, \text{s} $ is $ 6 \, \text{m} $.

### 2. What distance was traveled from $ t = 3 \, \text{s} $ to $ 5 \, \text{s} $?
# Explanation:
## Step 1: Find positions at $ t = 3 \, \text{s} $ and $ t = 5 \, \text{s} $
From the graph, at $ t = 3 \, \text{s} $, the position ($ x_1 $) is $ 3 \, \text{m} $. At $ t = 5 \, \text{s} $, the position ($ x_2 $) is $ 8 \, \text{m} $ (assuming the graph's scale, the vertical jump from $ t = 3 $ to $ t = 5 $ shows this change).
## Step 2: Calculate distance
Distance is the absolute difference between the final and initial positions. So, distance $ d=\vert x_2 - x_1\vert=\vert8 - 3\vert = 5 \, \text{m} $.
# Answer:
The distance traveled from $ t = 3 \, \text{s} $ to $ t = 5 \, \text{s} $ is $ 5 \, \text{m} $.

### 3. During what time intervals was there no motion?
# Explanation:
## Step 1: Recall the meaning of no motion on a position - time graph
On a position - time graph, no motion (zero velocity) is represented by a horizontal line (since position does not change with time).
## Step 2: Identify the horizontal segments
- From $ t = 0 \, \text{s} $ to $ t = 3 \, \text{s} $: The position is constant (horizontal line), so no motion.
- From $ t = 5 \, \text{s} $ to $ t = 8 \, \text{s} $: The position is constant (horizontal line), so no motion.
- From $ t = 15 \, \text{s} $ to $ t = 17 \, \text{s} $: The position is constant (horizontal line), so no motion.
# Answer:
The time intervals with no motion are $ 0 \, \text{s}-3 \, \text{s} $, $ 5 \, \text{s}-8 \, \text{s} $, and $ 15 \, \text{s}-17 \, \text{s} $.

### 4. During which time interval was motion the slowest?
# Explanation:
## Step 1: Recall the relationship between slope and speed on a position - time graph
The speed is the magnitude of the slope of the position - time graph. A smaller slope (closer to zero) means slower motion.
## Step 2: Analyze the slopes of each interval with motion
- From $ t = 3 \, \text{s} $ to $ t = 5 \, \text{s} $: The slope is large (steep line), so speed is high.
- From $ t = 8 \, \text{s} $ to $ t = 15 \, \text{s} $: The slope is moderate.
- From $ t = 17 \, \text{s} $ to $ t = 20 \, \text{s} $: Let's calculate the slope. The change in position ($ \Delta x $) from $ t = 17 \, \text{s} $ (position $ 3 \, \text{m} $) to $ t = 20 \, \text{s} $ (position $ 0 \, \text{m} $) is $ 0 - 3=- 3 \, \text{m} $, and change in time ($ \Delta t $) is $ 20 - 17 = 3 \, \text{s} $. The slope is $ \frac{\Delta x}{\Delta t}=\frac{-3}{3}=- 1 \, \text{m/s} $ (magnitude 1 m/s).
- Now, compare with other intervals. Wait, maybe I made a mistake. Wait, let's re - check. Wait, from $ t = 8 \, \text{s} $ to $ t = 15 \, \text{s} $: Let's say at $ t = 8 \, \text{s} $ position is $ 8 \, \text{m} $, at $ t = 15 \, \text{s} $ position is $ 3 \, \text{m} $. $ \Delta x=3 - 8=- 5 \, \text{m} $, $ \Delta t = 15 - 8 = 7 \, \text{s} $. The magnitude of the slope is $ \frac{5}{7}\approx0.71 \, \text{m/s} $. Wait, no, maybe I misread the graph. Wait, actually, the interval from $ t = 17 \, \text{s} $ to $ t = 20 \, \text{s} $: position at $ t = 17 $ is $ 3 \, \text{m} $, at $ t = 20 $ is $ 0 \, \text{m} $. $ \Delta x=- 3 \, \text{m} $, $ \Delta t = 3 \, \text{s} $, slope magnitude $ 1 \, \text{m/s} $. From $ t = 8 $ to $ t = 15 $: $ \Delta x=3 - 8=- 5 \, \text{m} $, $ \Delta t = 7 \, \text{s} $, slope magnitude $ \frac{5}{7}\approx0.71 \, \text{m/s} $. Wait, but maybe the graph has different markings. Wait, alternatively, the interval with the least steep slope (smallest speed) is $ 17 \, \text{s}-20 \, \text{s} $? Wait, no, maybe I messed up. Wait, another approach: the time interval where the object moves the least distance in a given time. Wait, from $ t = 17 $ to $ t = 20 $, time elapsed is 3 s, distance moved is 3 m. From $ t = 8 $ to $ t = 15 $, time elapsed is 7 s, distance moved is 5 m (from 8 m to 3 m). The speed in $ 8 - 15 $ is $ \frac{5}{7}\approx0.71 \, \text{m/s} $, in $ 17 - 20 $ is $ \frac{3}{3}=1 \, \text{m/s} $. Wait, maybe the correct interval is $ 8 \, \text{s}-15 \, \text{s} $? Wait, no, maybe the graph's scale is different. Wait, perhaps the slowest is $ 17 \, \text{s}-20 \, \text{s} $? Wait, no, let's check again. Wait, the key is that the slope (speed) is smallest when the line is least steep. Looking at the graph, the segment from $ t = 17 \, \text{s} $ to $ t = 20 \, \text{s} $ has a slope, and the segment from $ t = 8 \, \text{s} $ to $ t = 15 \, \text{s} $ has a more gentle slope? Wait, maybe I made a mistake in calculation. Let's assume the graph: at $ t = 8 $, position is 8 m; at $ t = 15 $, position is 3 m. So $ \Delta x = 3 - 8=- 5 $, $ \Delta t=15 - 8 = 7 $, speed $ \vert\frac{-5}{7}\vert\approx0.71 \, \text{m/s} $. At $ t = 17 $, position is 3 m; at $ t = 20 $, position is 0 m. $ \Delta x=0 - 3=- 3 $, $ \Delta t = 3 $, speed $ \vert\frac{-3}{3}\vert = 1 \, \text{m/s} $. At $ t = 3 $ to $ t = 5 $: $ \Delta x=8 - 3 = 5 $, $ \Delta t = 2 $, speed $ \frac{5}{2}=2.5 \, \text{m/s} $. So the slowest is $ 8 \, \text{s}-15 \, \text{s} $ because its speed is approximately $ 0.71 \, \text{m/s} $, which is less than 1 m/s and 2.5 m/s. Wait, but maybe the graph is different. Alternatively, if we consider that the slope from $ t = 8 $ to $ t = 15 $ is the least steep, so that's the slowest.
# Answer:
The motion was slowest during the time interval $ 8 \, \text{s}-15 \, \text{s} $ (or depending on the exact graph scale, but based on slope analysis, this interval has the smallest slope magnitude).

### 5. What distance was traveled from $ t = 17 \, \text{s} $ to $ 20 \, \text{s} $?
# Explanation:
## Step 1: Find positions at $ t = 17 \, \text{s} $ and $ t = 20 \, \text{s} $
From the graph, at $ t = 17 \, \text{s} $, the position ($ x_1 $) is $ 3 \, \text{m} $. At $ t = 20 \, \text{s} $, the position ($ x_2 $) is $ 0 \, \text{m} $.
## Step 2: Calculate distance
Distance is the absolute difference between the final and initial positions. So, distance $ d=\vert x_2 - x_1\vert=\vert0 - 3\vert = 3 \, \text{m} $.
# Answer:
The distance traveled from $ t = 17 \, \text{s} $ to $ t = 20 \, \text{s} $ is $ 3 \, \text{m} $.

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QUESTION IMAGE

Question

Explanation:

1. What is the position at \( t = 11 \, \text{s} \)?

Step 1: Analyze the graph

Step 2: Read the position

Step 1: Find positions at \( t = 3 \, \text{s} \) and \( t = 5 \, \text{s} \)

Step 2: Calculate distance

Step 1: Recall the meaning of no motion on a position - time graph

Step 2: Identify the horizontal segments

Answer:

2. What distance was traveled from \( t = 3 \, \text{s} \) to \( 5 \, \text{s} \)?